Calculating deflections in cracked state with CEDRUS according to Eurocode 2

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Due to a query we got from a client regarding the calculation of deflections in cracked state with CEDRUS, in which the results from calculus programs (CEDRUS among them) were compared with the ones obtained manually, we decided to carry out a comparative analysis between the values obtained with a manual calculations following the methods of the Eurocode 2 and the ones obtained with CEDRUS.

The method adopted by the EC2 is based on the calculation of the curvature of the cross sections under bending. The deflections are obtained from those curvatures. A rigorous calculation of the deflections would be to calculate the curvature of different intervals all along the element and then use numerical integration techniques to estimate the critical deflections. This should be done considering that some of the cross sections will be cracked under load and some other, in regions with a lower moment, will be uncracked. This laborious process is rarely justified, that is why the following procedure, based on the EC2 and being the one used in CEDRUS, assumes that it is accurate enough to calculate the curvature of the beam or slab considering both the cracked and the uncracked cross sections and using an ‘average’ value to estimate the final deflection from the standard formula for deflections, based on elastic theory.

- CALCULATING THE CURVATURE

The curvature is calculated using both the cracked and the uncracked cross sections. An estimation of the ‘average’ value of the curvature is obtained with the following formula:

 

Where:

1/r = average curvature

= values of the curvature calculated for cracked and uncracked cases, respectively.

= coefficient given by .

= load duration factor (1 for short term load; 0.5 for sustained or cyclic loads). In CEDRUS it takes a value of 0.5.

 = stress on the steel in tension, calculated for the cracked cross section.

= stress on the steel in tension, calculated over the cracked cross section under the load that exactly causes the cracking in the section.

For the calculation of , the ratio  can be obtained in a more convenient way using , where the first term is the cracking moment and the second term is the design moment for the calculation of the curvature and the deflection.

If we take the simple example of a solid slab of 8 metres in length, 1 metre in width and 0.30 metres in thickness, doubly-supported on two walls that are infinitely stiff in the vertical direction (z) and twist-restraint free with respect to the ‘x’ and ‘y’ axes:

 

The concrete used for the slab is HA-25 class, with tensile strength to 28 days and elastic modulus of E_cm=32 GPa.

Therefore, with this data, a cracking moment and a gross inertia of the cross section are obtained.

The deflection in cracked state and the gross elastic deflection for 3 different calculation hypotheses must be calculated. These 3 hypotheses are explained below and their purpose is to analyse the behaviour of a completely cracked model in different situations:

• Case 1: Low moment and low quantity: M1 = 1.5 Mcrk (As1)
• Case 2: High moment and high quantity: M2 = 4 Mcrk (As2)
• Case 3: Low moment and high quantity: M1 = 1.5 Mcrk (As2)

M₁ and M₂ are moments obtained for SLS and the bottom face reinforcement has a value of 12.0 cm²/m (A_s1,inf) in case 1 and of 37.2 cm²/m (A_s2,inf) in cases 2 and 3.

To avoid a variation on the bending moment for each cross section of the piece, it has been decided to impose the moments M₁ and M₂ directly on both ends.

Hence, the exposed data are shown in tabular form below. Notice that the inferior reinforcements introduced verify the ULS check, always having the minimum geometrical quantity at the top face.

- CALCULATING THE DEFLECTION FROM THE CURVATURE

The deflection of the element can be calculated from the curvature using the elastic theory for bending that, for small deflections, is based on the expression:

 

Where Mx is the bending moment in a cross section at a distance x from the origin according to the following figure. It shows a doubly-supported beam with a constant moment applied at both ends (we will use this conditions for the rest of this article).

 

For small deflections, the term  is approximately equal to the curvature, which is reciprocal to the radius of curvature. By integrating  we obtain an expression for the deflection. This will be particular for this specific case, doubly-supported with a constant moment M along the full span, thus, Mx = M.

Hence,

 

As the slope is zero at mid-span, where x = L/2, then:

 

And

 

Integrating again we obtain:

 

At support A, when x = 0, y = 0, resulting in D = 0 and therefore:

 at any cross section.

The maximum deflection is found at mid-span, where x = L/2, in which case:

Taking into account that for any uncracked cross section:

 

The maximum deflection can be expressed as:

 

Applying the stated theory and taking the data previously shown, the results we obtain in each of the cases are:

• CASE 1:

Curvature of the uncracked cross section:

 

Curvature of the cracked cross section:

 

Average of the curvatures:

 

So the elastic deflection is:

 

And the total deflection:

 

CEDRUS: 49.38 mm → (+ 1.06)

• CASE 2:

Curvature of the uncracked cross section:

 

Curvature of the cracked cross section:

 

Average of the curvatures:

 

So the elastic deflection is:

 

And the total deflection:

 

CEDRUS: 63.16 mm → (+ 1.00)

• CASE 3:

Curvature of the uncracked cross section:

 

Curvature of the cracked cross section:

 

Average of the curvatures:

 

So the elastic deflection is:

 

And the total deflection:

 

CEDRUS: 21.06 mm → (+ 1.01)

As we can observe, carrying out the method of the EC2, we practically get to the same values as CEDRUS, that is, the results that we obtain in CEDRUS correspond with the total deflection considering the distribution coefficients that the standard specifies.